32j^2-200=0

Solution for 32j^2-200=0 equation:

32j^2-200=0

a = 32; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·32·(-200)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160}{2*32}=\frac{-160}{64}
=-2+1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160}{2*32}=\frac{160}{64}
=2+1/2
$